Day 41 | 343. Integer Break | 96. Unique Binary Search Trees

Day 1 | 704. Binary Search | 27. Remove Element | 35. Search Insert Position | 34. First and Last Position of Element in Sorted Array
Day 2 | 977. Squares of a Sorted Array | 209. Minimum Size Subarray Sum | 59. Spiral Matrix II
Day 3 | 203. Remove Linked List Elements | 707. Design Linked List | 206. Reverse Linked List
Day 4 | 24. Swap Nodes in Pairs| 19. Remove Nth Node From End of List| 160.Intersection of Two Lists
Day 6 | 242. Valid Anagram | 349. Intersection of Two Arrays | 202. Happy Numbe | 1. Two Sum
Day 7 | 454. 4Sum II | 383. Ransom Note | 15. 3Sum | 18. 4Sum
Day 8 | 344. Reverse String | 541. Reverse String II | 替换空格 | 151.Reverse Words in a String | 左旋转字符串
Day 9 | 28. Find the Index of the First Occurrence in a String | 459. Repeated Substring Pattern
Day 10 | 232. Implement Queue using Stacks | 225. Implement Stack using Queue
Day 11 | 20. Valid Parentheses | 1047. Remove All Adjacent Duplicates In String | 150. Evaluate RPN
Day 13 | 239. Sliding Window Maximum | 347. Top K Frequent Elements
Day 14 | 144.Binary Tree Preorder Traversal | 94.Binary Tree Inorder Traversal| 145.Binary Tree Postorder Traversal
Day 15 | 102. Binary Tree Level Order Traversal | 226. Invert Binary Tree | 101. Symmetric Tree
Day 16 | 104.MaximumDepth of BinaryTree| 111.MinimumDepth of BinaryTree| 222.CountComplete TreeNodes
Day 17 | 110. Balanced Binary Tree | 257. Binary Tree Paths | 404. Sum of Left Leaves
Day 18 | 513. Find Bottom Left Tree Value | 112. Path Sum | 105&106. Construct Binary Tree
Day 20 | 654. Maximum Binary Tree | 617. Merge Two Binary Trees | 700.Search in a Binary Search Tree
Day 21 | 530. Minimum Absolute Difference in BST | 501. Find Mode in Binary Search Tree | 236. Lowes
Day 22 | 235. Lowest Common Ancestor of a BST | 701. Insert into a BST | 450. Delete Node in a BST
Day 23 | 669. Trim a BST | 108. Convert Sorted Array to BST | 538. Convert BST to Greater Tree
Day 24 | 77. Combinations
Day 25 | 216. Combination Sum III | 17. Letter Combinations of a Phone Number
Day 27 | 39. Combination Sum | 40. Combination Sum II | 131. Palindrome Partitioning
Day 28 | 93. Restore IP Addresses | 78. Subsets | 90. Subsets II
Day 29 | 491. Non-decreasing Subsequences | 46. Permutations | 47. Permutations II
Day 30 | 332. Reconstruct Itinerary | 51. N-Queens | 37. Sudoku Solver
Day 31 | 455. Assign Cookies | 376. Wiggle Subsequence | 53. Maximum Subarray
Day 32 | 122. Best Time to Buy and Sell Stock II | 55. Jump Game | 45. Jump Game II
Day 34 | 1005. Maximize Sum Of Array After K Negations | 134. Gas Station | 135. Candy
Day 35 | 860. Lemonade Change | 406. Queue Reconstruction by Height | 452. Minimum Number of Arrows
Day 36 | 435. Non-overlapping Intervals | 763. Partition Labels | 56. Merge Intervals
Day 37 | 738. Monotone Increasing Digits | 714. Best Time to Buy and Sell Stock | 968. BT Camera
Day 38 | 509. Fibonacci Number | 70. Climbing Stairs | 746. Min Cost Climbing Stairs
Day 39 | 62. Unique Paths | 63. Unique Paths II

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LeetCode 343. Integer Break

Question Link

class Solution {
    public int integerBreak(int n) {
        // dp[i] is the maximum product of positive integer i split results
        int[] dp = new int[n+1];
        dp[2] = 1;
        for(int i = 3; i <= n; i++){
            for(int j = 1; j <= i/2; j++)
                // We have to add dp[i] into the comparison because the value dp[i] may become larger or smaller
                dp[i] = Math.max(dp[i], Math.max(j * dp[i - j], j * (i - j)));
        } 
        return dp[n];
    }
}

• Determine the meaning of dp[i]
  • dp[i] is the maximum product of positive integer i split results
• Determine the recursive formula
  • dp[i] = Math.max(dp[i], Math.max(j * dp[i - j], j * (i - j)));
• Initialize the dynamic programming array
  • dp[2] = 1
• The traverse order is front to back and from top to bottom

LeetCode 96. Unique Binary Search Trees

Question Link

class Solution {
    public int numTrees(int n) {
        // The number of unique BST composed of i different nodes is dp[i]
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;
        for(int i = 2; i <= n; i ++){
            for(int j = 1; j <= i; j++)
                // `j - 1` is the node number of the left subtree when `j` is the root node.
	            // `i - j` is the node number of the right subtree when `j` is the root node.
                dp[i] += dp[j - 1] * dp[i - j];
            
        }
        return dp[n];
    }
}

• Determine the meaning of dp[i]
  • The number of unique BST composed of i different nodes is dp[i]
• Determine the recursive formula
  • dp[i] += dp[j - 1] * dp[i - j];
  • j - 1 is the node number of the left subtree when j is the root node.
  • i - j is the node number of the right subtree when j is the root node.
• Initialize the dynamic programming array
  • dp[0] = 1;
  • dp[1] = 1;
• By traverse j, sum up the number of BST when j is the root node